Answer
(a) $f$ is decreasing on the interval $(0, 1)$
$f$ is increasing on the interval $(1,\infty)$
(b) The local minimum is $f(1) = 0$
There is no local maximum.
(c) The graph is concave up on this interval: $(0, \infty)$
There are no points of inflection.
Work Step by Step
(a) $f(x) = x^2-x-ln~x$
Note that this function is defined on the interval $(0, \infty)$
We can find the points where $f'(x) = 0$:
$f'(x) = 2x-1-\frac{1}{x} = 0$
$2x^2-x-1 = 0$
$(2x+1)(x-1) = 0$
$x = -\frac{1}{2}, 1$
When $0 \lt x \lt 1$ then $f'(x) \lt 0$
$f$ is decreasing on the interval $(0, 1)$
When $1 \lt x$ then $f'(x) \gt 0$
$f$ is increasing on the interval $(1,\infty)$
(b) $f(1) = (1)^2-(1)-ln(1) = 0$
The local minimum is $f(1) = 0$
(c) We can find the points where $f''(x) = 0$:
$f''(x) = 2+\frac{1}{x^2} = 0$
Since $(2+\frac{1}{x^2}) \gt 0$ for all values of $x$, there are no solutions for the equation
$f''(x) = 2+\frac{1}{x^2} = 0$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(0, \infty)$
There are no points of inflection.
